Ncert Solutions Physics Class 12th

6.14 Length of the rod, l = 15 cm = 0.15 m

Magnetic field strength, B = 0.5 T

Resistance of the closed loop, R = 9 mΩ = 9 $*{10}^{-3}$ Ω

Induced emf = 9 mV

Polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

Speed of the rod, v = 12 cm/s = 0.12 m/s

Induced emf is given as, e = Bvl = 0.5 $*0.12*0.15$ = 9 mV

Yes, when key K is closed, excess charge is maintained by the continuous flow of current. When the key K is open, there is excess charge built up at both ends of the rods.

Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod. There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

Retarding force exerted on the rod, F = IBl , where

I = Current flowing through the rod = $\frac{e}{R}$ = $\frac{9*{10}^{-3}}{9\mathrm{}*{10}^{-3}}$ = 1 A

F = 1 $*0.5*0.15$ = 0.075 N

Speed of the rod, v = 12 cm/s = 0.12 m/s

Power is given by P = Fv = 0.075 $*0.12=9*{10}^{-3}$ W = 9 mW

When key K is open, no power is expended.

Power dissipated by heat = $I^{2}R=1*$ 9 $*{10}^{-3}$ W = 9 mW

The source of this power is an external agent.

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.

6.13 Area of the coil, A = 2 ${cm}^2$ = 2 $*{10}^{-4}$ $m^2$

Number of turns, n = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 $*{10}^{-3}$ C

Total resistance of the coil and galvanometer, R = 0.50 Ω

We know, Induced current in the coil, I = $\frac{Inducedemf\left(e\right)}{R}$ ……… (1)

Induced emf is given by the equation, e = $-$ n $\frac{d\varnothing }{dt}$ ………………… (2),

Where d $\varnothing$ is the change of flux

Combining equations (1) and (2), we get

I = $\frac{-\mathrm{n}\frac{d\varnothing }{dt}}{R}$ or Idt = $-\frac{n}{R}$ d $\varnothing$ ……… (3)

Initial flux through coil, ${\varnothing }_i$ = BA, where B = Magnetic field strength

Final flux through coil, ${\varnothing }_f$ = 0

Integrating equation (3), we get

$\int Idt$ = $-\frac{n}{R}{\int }_{{\varnothing }_i}^{{\varnothing }_f}d\varnothing$

$\int Idt=Q=\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{f}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{i}\mathrm{l}$

Therefore

Q = $-\frac{n}{R}\left({\varnothing }_f-{\varnothing }_i\right)=-\frac{n}{R}\left(0-{\varnothing }_i\right)=\frac{n{\varnothing }_i}{R}=\frac{nBA}{R}$

B = $\frac{QR}{nA}$ = $\frac{7.5\mathrm{}*{10}^{-3}*0.5}{25*2*{10}^{-4}}$ = 0.75 T

Hence, the field strength is 0.75 T

6.12 Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 $*0.12=0.0144m^2$

Velocity of the loop, v = 8 cm /s = 0.08 m/s

Gradient of the magnetic field along negative x – direction.

$\frac{dB}{dx}$ = ${10}^{-3}$ T/cm = ${10}^{-1}$ T/m

Rate of decrease of magnetic field

$\frac{dB}{dt}$ = ${10}^{-3}$ T/s

Resistance of the loop, R = 4.50 mΩ = 4.5 $*{10}^{-3}$ Ω

Rate of change of the magnetic flux due to motion of the loop in a non-uniform magnetic field is given as:

$\frac{d\varnothing }{dt}$ = A $*\frac{dB}{dx}$ $*v$ = $0.0144*{10}^{-1}*0.08$ = 1.152 $*{10}^{-4}$ T $m^2{s}^{-1}$

Rate of change of flux due to explicit time variation in field B is given as:

$\frac{d\varnothing \text{'}}{dt}$ = A $*\frac{dB}{dt}$ = $0.0144*{10}^{-3}$ = 1.44 $*{10}^{-5}$ T $m^2{s}^{-1}$

Since the rate of change of the flux is the induced emf, the total emf in the loop can be calculated as:

e = 1.152 $*{10}^{-4}$ + 1.44 $*{10}^{-5}$ = 1.296 $*{10}^{-4}$ V

The induced current, I = $\frac{e}{R}$ = $\frac{1.296\mathrm{}*{10}^{-4}}{4.5\mathrm{}*{10}^{-3}}$ = 28.8 $*{10}^{-3}$ A

Therefore, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

6.11 The area of the rectangular coil, A = 8 $*2=$ 16 ${cm}^2$ = 16 $*{10}^{-4}$ $m^2$

Initial value of the magnetic field, $B_1$ = 0.3 T

Rate of decrease of the magnetic field, $\frac{dB}{dt}$ = 0.02 T/s

From the relation of induced emf e = $\frac{d\varnothing }{dt}$ , where $d\varnothing$ is the change in the flux linkage with the coil = A $*B$

Hence, e = $\frac{d\left(AB\right)}{dt}$ = A $\frac{dB}{dt}$ = 16 $*{10}^{-4}*0.02$ = 3.2 $*{10}^{-5}$ V

Resistance in the loop, R = 1.6 Ω

Hence I = $\frac{e}{R}$ = $\frac{3.2\mathrm{}*{10}^{-5}}{1.6}$ = 2 $*{10}^{-5}$ A

Power dissipated in the form of heat is given by

P = $I^2$ R = ( ${\mathrm{}2\mathrm{}*{10}^{-5})}^2*1.6$ = 6.4 $*{10}^{-10}$ W

The source of heat loss is an external agent, which is responsible for changing the magnetic field with time.

2.36 We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

Yes, the person will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

The occurrence of thunderstorm and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.

During lightning and thunderstorm, light energy, heat energy and sound enegy are dissipated in the atmosphere.